Joakim Edsj¨ o

Joakim Edsj¨ o

Fysikum, Stockholms Universitet Tel: 08-674 76 48

Solutions to

Exam in Analytical Mechanics, 5p

June 2, 2001

Solutions will eventually also be available on

http://www.physto.se/~edsjo/teaching/am/index.html . Problem 1

a) The kinetic energy is given by T = 1

2 m

R sin θω ϕ + R ˙θˆ ˆ θ  ₂

= 1 2 mR ²

ω ² sin ² θ + ˙ θ ²



and the potential energy is given by

U = mgR (1 − cos θ) The Lagrangian is then given by

L = T − U = 1 2 mR ²

ω ² sin ² θ + ˙ θ ²

 − mgR (1 − cos θ)

and it’s derivatives are

 _∂L

∂θ = mR ² ω ² sin θ cos θ − mgR sin θ

∂L

∂ ˙θ = mR ² θ ˙ Inserted into Lagrange’s equations,

d dt

 ∂L

∂ ˙ θ



− ∂L

∂θ = 0 we get

mR ² θ = mR ¨ ² ω ² sin θ cos θ − mgR sin θ (1) which is the equation of motion for θ.

b) From Eq. (1) we see that ¨ θ = 0 for θ = 0 and hence θ = 0 is an equilibrium point. To find if it is stable or not, we Taylor expand the right-hand side in Eq. (1) keeping terms up to linear order in θ, i.e. we set 

sin θ  θ cos θ  1 which gives

mR ² θ ¨  

mR ² ω ² − mgR 

θ

This equation has oscillating cos and sin solutions if the coefficient in front of θ in the right- hand side is negative, otherwise the solution is exponentials. Hence, for the solution to be stable, the coefficient has to be negative, i.e.

mR ² ω ² − mgR < 0

⇒ ω ² < _R ^g

⇒ ω c = _g

R

c) From Eq. (1) we see that ¨ θ = 0 when sin θ 

mR ² ω ² cos θ − mgR 

= 0.

We see that this equation is fulfilled when

sin θ = 0 or cos θ = g Rω ²

The first of these gives the two equilibrium points θ = 0 and θ = π, whereas the second equation only has a solution when ω > ω _c and then the equilibrium point is

θ = arccos g Rω ²

One can show that this equilibrium point is stable by Taylor expand the right-hand side in Eq. (1) around θ = arccos _Rω ^g

.

Uppgift 2

a) Choose z as the height for the mass m as the generalized coordinate. The kinetic energy for the mass m is then given by

T m = 1 2 m ˙z ²

When the mass m has moved a distance z, the mass M has moved the distance z/2. It’s kinetic energy gets a contribution both from the translation of the centre of mass, but also from the rotation around the centre of mass,

α

m M radius R

A z

T _M = 1 2 M

 1 2 z ˙

 ₂ + 1

2 Iω ²

For a solid cylinder the moment of inertia for rotation around the symmetry axis is given by I = ¹ ₂ M R ² . The angular velocity is given by ω = _2R ^˙z . The kinetic energy for M is then given by

T _M = 1

8 M ˙ z ² + 1 2 1 2 M R ²

 z ˙ 2R

 ₂

= 1

8 M ˙ z ² + 1

16 M ˙ z ² = 3 16 M ˙ z ² . The potential energy is given by

U = mgz − Mg 1

2 z sin α

which finally gives us the Lagrangian L = T _m + T _M − U =

 1 2 m + 3

16 M



˙ z ² −

 mg − 1

2 M g sin α

 z.

It’s derivatives are  _∂L

∂z = ¹ ₂ M g sin α − mg

∂L ∂ ˙z = 

m + ³ ₈ M 

˙z Lagrange’s equations,

d dt

 ∂L

∂ ˙ z



− ∂L

∂z = 0

then yield 

m + 3 8 M



¨ z = 1

2 M g sin α − mg (2)

which is easily integrated to give the solution z(t) = 2M g sin α − 4mg

8m + 3M t ² + At + B ; A, B = constants b) The system is in equilibrium when ¨ z = 0. Eq. (2) tell us that ¨ z = 0 when

1 2 M g sin α − mg = 0

⇒ α = arcsin ^2m _M .

We also see that equilibrium only can be obtained when M ≥ 2m.

Uppgift 3

a) Vi have that {f, gh} =

i

 ∂f

∂p _i

∂(gh)

∂q _i − ∂f

∂q _i

∂(gh)

∂p _i 

=

i

 ∂f

∂p i

∂g

∂q i h + g ∂f

∂p i

∂h

∂q i − ∂f

∂q i

∂g

∂p i h − g ∂f

∂q i

∂h

∂p i

= g

i

 ∂f

∂p i

∂h

∂q i − ∂f

∂q i

∂h

∂p i

+

i

 ∂f

∂p i

∂g

∂q i − ∂f

∂q i

∂g

∂p i

h = g {f, h} + {f, g}h.

The second relation is easily shown in the same way.

b) To determine the condition β and γ has to fulfill for L z to be a constant of motion, we can e.g. use Noether’s theorem. Alternatively, we can use Poisson brackets by noting that

dL _z

dt = {H, L z } + ∂L _z  ∂t

0

= {H, L z }.

In other words, we want to determine β and γ such that {H, L z } = 0. The Hamiltonian is given by

H = 1 2m

 p ² _x + p ² _y + p ² _z 

+ αz ² e ^βx

^+γy

The z component of the angular momentum is given by L z = xp y − yp x .

We are now ready to calculate the Poisson bracket between H and L z . We then use the relations that were proved in a) and that {q i , q j } = 0, {p i , p j } = 0, {p i , q j } = δ ij and {q i , p j } = −δ ij to simplify our expression,

{H, L z } = { 1 2m

 p ² _x + p ² _y + p ² _z 

+ αz ² e ^βx

^+γy

, x p _y − yp x } =

= 1

2m {p ² x , x p _y } − 1

2m {p ² y , yp _x } + {αz ² e ^βx

^+γy

, x p _y } − {αz ² e ^βx

^+γy

, yp _x }

= 1

2m p _x {p   x , x }

1

p _y − 1

2m p _y {p y , y }  

1

p _x + αz ² x {e ^βx

^+γy

, p _y } − αz ² y {e ^βx

^+γy

, p _x }

= 1

2m [p _x p _y − p y p _x ]

 

0

+αz ² x

i



 

 

∂e ^βx

^+γy

∂p _i

∂p _y

∂q _i  0

− ∂e ^βx

^+γy

∂q _i

∂p _y

∂p _i  δ



 

 

−αz ² y

i



 

 

∂e ^βx

^+γy

∂p _i

∂p x

∂q _i  0

− ∂e ^βx

^+γy

∂q _i

∂p x

∂p _i  δ



 

 

= −αz ² x ∂e ^βx

^+γy

∂y + αz ² y ∂e ^βx

^+γy

∂x

= −αz ² x2γye ^βx

^+γy

+ αz ² y2βxe ^βx

^+γy

= αz ² xye ^βx

^+γy

[β − γ]

⇒ {H, L z } = 0 if β = γ

L _z is thus conserved if β = γ, which is the condition we looked for.

Remark. {H, L z } = 0 is also fulfilled if x = 0, y = 0 or z = 0, but if {H, L z } = 0 should be fulfilled for arbitrary initial conditions we have to have β = γ.

Uppgift 4

a) See Scheck, section 2.5 or the lecture notes.

b) This is easy to show with calculus of variations. The distance between (x ₀ , y ₀ ) and (x ₁ , y ₁ ) is given by

L =

 x

x

ds =

 x

x

1 + y
² dx

We can use Euler’s equation given in 4a with f (y, y
, x) =

1 + y
² Inserted into Euler’s equation, we get

0 = d dx

 ∂f

∂y



− ∂f

∂y = d dx

 y
1 + y
²



− 0 = 0

which is easily integrated to y

1 + y
² = A = const. ⇒ y
= B = const.

Integrating once more, we get

y = Bx + C

which is the equation for the straight line. The constants B and C are given by the condition that the line has to pass through (x ₀ , y ₀ ) and (x ₁ , y ₁ ).

Uppgift 5

a) For a generating function of the type U we have that q i = − ∂U

∂p _i ; P j = − ∂U

∂Q _j ; H = H + ˜ ∂U

∂t . We want the new Hamiltonian, ˜ H, to be identically equal to zero, i.e. that

H(q

, p

, t) + ∂U

∂t = 0 Use that q _i = − _∂p ^∂U

and we get

H( − ∂U

∂p _i

, p

, t) + ∂U

∂t = 0

which is the Hamilton-Jacobi equation in the momentum representation. This is a partial differential equation for U with respect to p

and t.

b) With the given Hamiltonian, Hamilton-Jacobi’s equation in the momentum representation yield

p ²

2m − mg ∂U

∂p + ∂U

∂t = 0 (3)

Make the Ansatz

U (p, t) = U ₁ (p) + U ₂ (t)

(where the dependence on the constant Q is not explicitly given). Inserted into Eq. (3) we get

p ²

2m − mg ∂U ₁  ∂p 

=E

+ ∂U ₂  ∂t

=−E

= 0

where we realize that the first two terms only depend on p whereas the last term only depend on t and thus they have to be equal constants (but with opposite sign). We then get one equation for U ₁ and one for U ₂ ,

 _p

2m − mg ^∂U _∂p

= E

∂U

∂t = −E.

These are easily solved and we get



U ₁ = _6m ^p

_g − ^Ep _mg + const.

U ₂ = −Et + const.

which yield the generating function U we looked for,

U = p ³

6m ² g − Ep

mg − Et + const. (4)

U should be a function of Q, p and t though so our separation constant E has to be a function of our constant Q. We choose to define E = Q and let the arbitrary constant in Eq. (4) be zero, which yield

U (Q, p, t) = p ³

6m ² g − Qp

mg − Qt. (5)

We can now get the equations that give us the canonical transformation,



q = − ^∂U _∂p = − _2m ^p

g + _mg ^Q P = − ^∂U _∂Q = _mg ^p + t The second of these equations gives

p = mg (P − t) (6)

which, if inserted into the first equation, gives

q = − g (P − t) ²

2 + Q

mg (7)

Hamilton’s canonical equations for the new variables Q and P are trivial,

 Q ˙ = ^{∂ ˜} _∂P ^H = 0

P ˙ = − ^{∂ ˜} _∂Q ^H = 0 ⇒

 Q = β = const.

P = α = const.

Inserted into Eqs. (6)–(7) we get the solution



q(t) = _mg ^β − ^g(α−t) ₂

p(t) = mg (α − t)

The initial condition p(0) = mv ₀ gives α = v ₀ /g and q(0) = 0 gives β = mv ² ₀ /2. The solution with the given initial conditions are thus

 



q(t) = ^v _2g

− ^g ₂

v

g − t  ₂ p(t) = mg

v

g − t